题目链接
题意:有N个点,下标从0开始,问的是有多少个子树满足给出的H+1个条件,H代表深度(从0开始),每个深度要求对应的结点的个数是大于等于ai的,满足其一就要加入贡献。
思路:直接Dsu,但是由于重儿子上面的贡献会被直接储存,所以求重儿子的时候,我们需要的是直接加上他的答案即可。
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
#include <unordered_map>
#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 5e5 + 7;
int N, H, head[maxN], cnt, need[maxN];
ll ans = 0;
struct Eddge
{
int nex, to;
Eddge(int a=-1, int b=0):nex(a), to(b) {}
}edge[maxN];
inline void addEddge(int u, int v)
{
edge[cnt] = Eddge(head[u], v);
head[u] = cnt++;
}
int siz[maxN], dfn[maxN], tot, rid[maxN], Wson[maxN], deep[maxN] = {0};
void pre_dfs(int u)
{
siz[u] = 1; dfn[u] = ++tot; rid[tot] = u; Wson[u] = -1;
int maxx = 0;
for(int i=head[u], v; ~i; i=edge[i].nex)
{
v = edge[i].to;
deep[v] = deep[u] + 1;
pre_dfs(v);
siz[u] += siz[v];
if(maxx < siz[v])
{
maxx = siz[v];
Wson[u] = v;
}
}
}
int now_have[maxN] = {0}, _get[maxN] = {0};
void dfs(int u, bool keep)
{
for(int i=head[u], v; ~i; i=edge[i].nex)
{
v = edge[i].to;
if(v == Wson[u]) continue;
dfs(v, false);
}
if(~Wson[u])
{
dfs(Wson[u], true);
_get[u] += _get[Wson[u]];
ans += _get[Wson[u]];
}
for(int i=head[u], v, id; ~i; i=edge[i].nex)
{
v = edge[i].to;
if(v == Wson[u]) continue;
for(int j=dfn[v]; j < dfn[v] + siz[v]; j++)
{
id = rid[j];
now_have[deep[id]]++;
if(now_have[deep[id]] == need[deep[id]])
{
ans ++;
_get[u]++;
}
}
}
now_have[deep[u]] ++;
if(now_have[deep[u]] == need[deep[u]]) { ans ++; _get[u]++; }
if(!keep)
{
_get[u] = 0;
for(int i=dfn[u], v; i<dfn[u] + siz[u]; i++)
{
v = rid[i];
now_have[deep[v]] = 0;
}
}
}
inline void init()
{
cnt = 0;
for(int i=0; i<N; i++) head[i] = -1;
}
int main()
{
scanf("%d%d", &N, &H);
init();
for(int i=1, ff; i<N; i++)
{
scanf("%d", &ff);
addEddge(ff, i);
}
for(int i=0; i<=H; i++)
{
scanf("%d", &need[i]);
if(!need[i]) ans += N;
}
pre_dfs(0);
dfs(0, false);
printf("%lld\n", ans);
return 0;
}